0.34 Phy1340: angular momentum -- rotational equilibrium  (Page 6/6)

For horizontal equilibrium

F4 = F2 = 26133 newtons

For vertical equilibrium

F3 = F1 = M*g = 2000kg*9.8m/s^2, or

F3 = 19600 newtons

The magnitude of the force = sqrt(F4^2 + F3^2), or

Magnitude = sqrt(26133^2 + 19600^2), or

Magnitude = 32666 newtons

Therefore, the magnitude of the force is 32666 newtons.

The angle of the force is

angle = atan(19600 newtons/26133 newtons) in degrees

angle = 36.9 degrees north of east

A ladder scenario

A ladder is sitting on a horizontal floor and leaning against a smooth wall. A bucket of paint is hanging from a rung 80-percent up from the bottom of theladder. The parameters are as follows:

• Length of ladder = L = 5 m
• Angle between ladder and floor = A = 50 degrees
• Mass of the bucket of paint = m = 2.5 kg
• Mass of the ladder = M = 12 kg
• Coefficient of static friction = u = ?

Part 1

Draw a vector diagram of the forces acting on the ladder.

Solution:

There are five forces acting on the ladder:

1. A force directed outward from the wall at the top of the ladder. Label this force F1.
2. A force pushing straight down due to the mass of the bucket of paint. Label this force F2.
3. A force pushing straight down due to the mass of the ladder. Label this force F3.
4. A force at the bottom of the ladder directed toward the wall due to static friction. Label this force F4.
5. A force at the bottom of the ladder directed straight up due to the weight of the ladder and the bucket of paint. Label this force F5.

Part 2

The ladder is in static equilibrium.

The coefficient of static friction between the ladder and the wall is zero.

What is the minimum possible value for the coefficient of static friction between the ladder and the floor.

F1 = ?

F2 = m*g = 2.5kg * 9.8m/s^2 = 24.5 newtons

F3 = M*g = 12kg*9.8m/s^2 = 117.6 newtons

F4 = F5*u

For vertical equilibrium,

F5 = F2 + F3 = (2.5kg + 12kg)*9.8m/s^2 = 142 newtons

For rotational equilibrium, the sum of the torques about the bottom of the ladder must be zero.

Compute the horizontal distance to the line of action of the weight of the ladder.

X1 = (L/2)*cos(50 degrees), or

X1 = ((5/2)m)*cos(50 degrees), or

X1 = 1.61m

Compute the horizontal distance to the line of action of the weight of the bucket of paint.

X2 = 0.8*5m*cos(50 degrees), or

X2 = 2.57m

Compute the vertical distance to the line of action of the horizontal force at the top of the ladder.

Y1 = 5m*sin(50 degrees), or

Y1 = 3.83m

Compute sum of the torques about the bottom of the ladder.

Y1*F1 - X2*F2 - X1*F3 = 0, or

F1 = (X2*F2 + X1*F3)/Y1, or

F1 = (2.57m*24.5 newtons+ 1.61m*117.6 newtons)/3.83m

F1 = 65.87 newtons

For horizontal equilibrium,

F4 = 65.87 newtons

By substitution,

F4 = F5*u = 65.87 newtons

u = (65.87 newtons)/F5, or

u = (65.87 newtons)/142 newtons, or

u = 0.46

While the coefficient of static friction could be higher than this and still achieve static equilibrium, if the coefficient were any lower, the ladder wouldslide away from the wall.

Repeat the computations

I encourage you to repeat the computations that I have presented in this lesson to confirm that you get the same results. Experiment withthe scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

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